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S-i-1-i-n-n-R-Can-S-be-solved-




Question Number 4668 by FilupSmith last updated on 20/Feb/16
S=Σ_(i=1) ^∞ ^i (√n),   n∈R  Can S be solved?
S=i=1in,nRCanSbesolved?
Answered by Yozzii last updated on 20/Feb/16
S(N)=Σ_(i=1) ^N n^(0.5i) =(√n)Σ_(i=1) ^N (n^(0.5) )^(i−1)   S(N)=(((√n)(((√n))^N −1))/( (√n)−1))  S(N)=((((√n))^(N+1) −(√n))/( (√n)−1))  (1) For n≥1⇒(√n)≥1⇒((√n))^(N+1) →∞ as   N→∞.  lim_(N→∞) S(N)=((((√n))^∞ −(√n))/( (√n)−1))=∞.  S(∞) does not exist.     (2)If 0≤n<1⇒0≤(√n)<1  ⇒((√n))^(N+1) →0 as N→∞.  ∴lim_(N→∞) S(N)=((((√n))^∞ −(√n))/( (√n)−1))=((√n)/(1−(√n)))  S(∞) exists.    (3)For −1<n<0 let n=−a, 0<a<1.  ⇒(√n)=(√(−a))=i(√a)=(√a)(cos0.5π+isin0.5)=e^(0.5iπ) (√a)  ⇒((√n))^(N+1) =e^(0.5iπ(N+1)) ((√a))^(N+1)   Since 0<a<1⇒((√a))^(N+1) →0 as N→∞.  ∴lim_(N→∞) S(N)=((0×e^(∞i) −(√n))/( (√n)−1))=((√n)/(1−(√n)))  S(∞)=((i(√a))/(1−i(√a)))=((i(√a)(1+i(√a)))/(1+a))  S(∞)=((−a+i(√a))/(1+a))=((n+i(√(−n)))/(1−n)).⇒S(∞)∈C.    (4) For n≤−1⇒((√n))^(N+1) →e^(i∞)  if n=−1.  ⇒a=1.  S(N)=((((√n))^(N+1) −(√n))/( (√n)−1))  =(((i(√a))^(N+1) −i(√a))/(i(√a)−1))  =((i(√a)((i(√a))^N −1))/(i(√a)−1))  =((i(√a)(1−i(√a))(1−(i(√a))^N ))/(1+a))  =(((a+i(√a))(1−(i(√a))^N ))/(1+a))  =((a−a(i(√a))^N +i(√a)−(i(√a))^(N+1) )/(1+a))  =((a+i(√a))/(1+a))−(((i(√a))^N (a+i(√a)))/(1+a))  S(N)=((a+i(√a))/(1+a))(1−a^(0.5N) e^(0.5πiN) )  lim_(N→∞) S(N)=((1+i)/2)(1−e^(i∞) ) which is undefined  since e^(i∞)  is indeterminate.    (5)If a>1 then lim_(N→∞) S(N)=((a+i(√a))/(1+a))(1−∞e^(i∞) )  ∞e^(i∞)  is undefined⇒S(∞) is undefined  for n≤−1.
S(N)=Ni=1n0.5i=nNi=1(n0.5)i1S(N)=n((n)N1)n1S(N)=(n)N+1nn1(1)Forn1n1(n)N+1asN.limNS(N)=(n)nn1=.S()doesnotexist.(2)If0n<10n<1(n)N+10asN.limNS(N)=(n)nn1=n1nS()exists.(3)For1<n<0letn=a,0<a<1.n=a=ia=a(cos0.5π+isin0.5)=e0.5iπa(n)N+1=e0.5iπ(N+1)(a)N+1Since0<a<1(a)N+10asN.limNS(N)=0×einn1=n1nS()=ia1ia=ia(1+ia)1+aS()=a+ia1+a=n+in1n.S()C.(4)Forn1(n)N+1eiifn=1.a=1.S(N)=(n)N+1nn1=(ia)N+1iaia1=ia((ia)N1)ia1=ia(1ia)(1(ia)N)1+a=(a+ia)(1(ia)N)1+a=aa(ia)N+ia(ia)N+11+a=a+ia1+a(ia)N(a+ia)1+aS(N)=a+ia1+a(1a0.5Ne0.5πiN)limNS(N)=1+i2(1ei)whichisundefinedsinceeiisindeterminate.(5)Ifa>1thenlimNS(N)=a+ia1+a(1ei)eiisundefinedS()isundefinedforn1.

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